\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 34, pp. 1--6.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2012/34\hfil Mean value theorem]
{Mean value theorem for holomorphic functions}
\author[D. \c{C}akmak, A. Tiryaki \hfil EJDE-2012/34\hfilneg]
{Devrim \c{C}akmak, Ayd\i n Tiryaki} % in alphabetical order
\address{Devrim \c{C}akmak \newline
Gazi University\\
Faculty of Education\\
Department of Mathematics Education\\
06500 Teknikokullar - Ankara, Turkey}
\email{dcakmak@gazi.edu.tr}
\address{Ayd\i n Tiryaki \newline
Izmir University\\
Faculty of Arts and Sciences\\
Department of Mathematics and Computer Sciences\\
35350 Uckuyular - Izmir, Turkey}
\email{aydin.tiryaki@izmir.edu.tr}
\thanks{Submitted November 29, 2011. Published February 29, 2012.}
\subjclass[2000]{39B22, 26A24}
\keywords{Rolle's theorem; mean value theorem; Flett's theorem;
\hfill\break\indent holomorphic function}
\begin{abstract}
This article presents a generalization of Myers' theorem and
when the boundary assumption $f'(a)=f'(b)$ is removed,
and to prove this result for holomorphic functions of one complex variable.
After that, the equivalence of Rolle's and mean value theorems in the complex
plane are proved.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks
\section{Introduction}
We know the following two results, usually covered in a first semester
calculus course, and used to solve a great variety of
problems in optimization, economics. etc.
Let $f$ be a continuous function on a closed interval $[a,b]$. The
difference between the values of $f$ at the endpoints of $[a,b]$,
if derivative $f'(a)$ exists, can be estimated by using $f'(a)$:
\begin{equation}
f(b)-f(a)\approx f'(a)(b-a), \label{equat0}
\end{equation}
where the approximation is good if $b-a$ is small. In fact, this is just the
tangent line approximation that takes the form
\begin{equation}
f(a+\Delta x)\approx f(a)+f'(a)\Delta x,
\end{equation}
with $\Delta x$ replaced by $b-a$. Actually, the approximation \eqref{equat0}
can be replaced by the exact formula
\begin{equation}
f(b)-f(a)=f'(c)(b-a),
\end{equation}
where the derivative $f'$ is evaluated at a suitable point $c$
between $a$ and $b$, rather than at the end point $a$. Here we assume that
$f$ is differentiable at every point between $a$ and $b$, and the choice of $c$
depends on the particular function $f$. This result, known as the mean value
theorem, is of great importance in mathematical analysis.
\begin{theorem}[Mean Value Theorem] \label{thmA}
Let $f$ be a real continuous function on $[a,b]$ and differentiable in $(a,b)$.
Then there exists a point $c\in (a,b)$ such that
\begin{equation}
f(b)-f(a)=f'(c)(b-a). \label{equat1}
\end{equation}
\end{theorem}
If $f(a)=f(b)$, then the mean value theorem reduces to Rolle's theorem which
is also the another most fundamental results in mathematical analysis.
\begin{theorem}[Rolle's Theorem] \label{thmB}
Let $f$ be a real continuous function on $[a,b]$ and differentiable in
$(a,b)$. Furthermore, assume $f(a)=f(b)$. Then there is a point
$c\in (a,b)$ such that $f'(c)=0$.
\end{theorem}
The equivalence between Rolle's and mean value theorems for real-valued functions
has been proved for example in \cite{Qazi}.
There are many other types of mean value theorems that are less known.
In particular, in 1958 Flett \cite{Flett} proved the following variation of
the mean value theorem.
\begin{theorem}[Flett's Theorem] \label{thmC}
Let $f:[a,b]\to \mathbb{R}$ be differentiable on $[a,b]$ and
$f'(a)=f'(b)$. Then there exists a point $c\in (a,b)$ such that
\begin{equation}
f(c)-f(a)=f'(c)(c-a).
\end{equation}
\end{theorem}
In 1977, Myers \cite{Myers} gave the following result which is a slight
modification of Flett's theorem.
\begin{theorem}[Myers' Theorem] \label{thmD}
Let $f:[a,b]\to\mathbb{R}$ be differentiable on $[a,b]$ and
$f'(a)=f'(b)$. Then there exists a point $c\in (a,b)$ such that
\begin{equation}
f(b)-f(c)=f'(c)(b-c).
\end{equation}
\end{theorem}
The geometric interpretation of above theorems can be found in
\cite{Flett,Myers,Silverman}. For other examples of mean value
theorems, we refer the reader to the references in this article.
In 1998, Sahoo and Riedel \cite{SahooRiedel} gave a generalization of
Flett's mean value theorem and removed the boundary assumption on the
derivatives of the function $f$.
\begin{theorem}[Sahoo and Riedel's Theorem] \label{thmE}
Let $f:[a,b]\to\mathbb{R}$ be differentiable on $[a,b]$. Then there exists a point
$c\in (a,b)$ such that
\begin{equation}
f(c)-f(a)=f'(c)(c-a)-\frac{1}{2}\frac{f'(b)-f'(a)}{b-a}(c-a)^{2}.
\end{equation}
\end{theorem}
In general, these results do not immediately extend to holomorphic functions
of one complex variable. For the case of Rolle's theorem, the function
$f(z)=e^{z}-1$ has value $0$ at $z=0$ and at $z=2\pi i$, but $f'(z)=e^{z}$ has no zeros
in the complex plane. Evard and Jafari \cite{EvardJafari} went around this difficulty
by working with the real and imaginary parts of a holomorphic function.
Another approach is taken by Samuelsson \cite{Samuelsson}.
Moreover, Flett's theorem is not valid for
complex-valued functions of one complex variable. To see this, consider the
function $f(z)=e^{z}-z$. Then $f$ is holomorphic, and $f'(z)=e^{z}-1$.
Therefore, we have $f'(2k\pi i)=e^{2k\pi i}-1=0$ for
all integers $k$. In particular, $f'(0)=f'(2\pi i)$, that
is, the derivatives of $f$ at the endpoints of the closed interval
$[0,2\pi i]$ are equal. Nevertheless, the equation $f(z)-f(0)=f'(z)z$ has no
solution on the interval $(0,2\pi i)$, as we now show. The equation above
gives $1-z=e^{-z}$ and, since $z=iy$, we obtain $1-iy=\cos y-i\sin y$. The
comparison of the real and imaginary parts gives the system $\cos y=1$ and $
\sin y=y$, which has no solution in the interval $(0,2\pi )$. Thus Flett's
theorem fails in the complex domain.
Now, we give some notation. Let $\mathbb{C}$ denote the set of complex numbers.
For distinct $a$ and $b$ in $\mathbb{C}$, let $[a,b]$ denote the set
$\{a+t(b-a)$: $t\in [ 0,1]\}$; we will
refer to $[a,b]$ as a line segment or a closed interval in $\mathbb{C}$.
Similarly, $(a,b)$ denotes the set $\{a+t(b-a)$ : $t\in (0,1)\}$.
In 1999, Davitt et al \cite{Davitt} prove a
version of Flett's theorem for holomorphic functions of one complex variable.
They gave a generalization of Theorem \ref{thmE} for holomorphic functions where
\begin{equation}
\langle u,v\rangle=\operatorname{Re}(u\overline{v})
\end{equation}
for any two complex numbers $u$ and $v$.
\begin{theorem}[Davitt, Powers, Riedel and Sahoo's Theorem] \label{thmF}
Let $f$ be a holomorphic function defined on an open convex subset $D_{f}$
of $\mathbb{C}$. Let $a$ and $b$ be two distinct points in
$D_{f}$. Then there exist $z_1,z_2\in (a,b)$ such that
\begin{equation}
\operatorname{Re}(f'(z_1))
=\frac{\langle b-a, f(z_1)-f(a)\rangle}{\langle b-a,z_1-a\rangle}+
\frac{1}{2}\frac{\operatorname{Re}(f'(b)-f'(a))}{b-a}(z_1-a)
\end{equation}
and
\begin{equation}
\operatorname{Im}(f'(z_2))=\frac{\langle b-a, -i [ f(z_2)-f(a)] \rangle}{
\langle b-a, z_2-a\rangle}+\frac{1}{2}\frac{\operatorname{Im}(f'(b)-f'(a))}{
b-a}(z_2-a).
\end{equation}
\end{theorem}
In this paper, our first aim is to present a generalization of Myers'
theorem and removed the boundary assumption on the derivatives of the
function $f$, i.e. $f'(a)=f'(b)$.
Our second aim is to provide this result for holomorphic functions of one complex variable.
After that the equivalence of Rolle's and mean value theorems in the complex plane are
proved.
\section{Main Results}
Our first goal of this paper is to extend Myers' theorem for real-valued
functions to a result that does not depend on the hypothesis $f'(a)=f'(b)$,
but reduces to Myers' theorem when this is the case.
\begin{theorem} \label{thm1}
If $f:[a,b]\to\mathbb{R}$ is a differentiable function, then there exists a
point $c\in (a,b)$ such that
\begin{equation}
f(b)-f(c)=f'(c)(b-c)+\frac{1}{2}\frac{f'(b)-f'(a)}{b-a}(b-c)^{2}.
\end{equation}
\end{theorem}
\begin{proof}
Consider the auxiliary function $h:[a,b]\to\mathbb{R}$ defined by
\begin{equation}
h(x)=f(x)-\frac{1}{2}\frac{f'(b)-f'(a)}{b-a}(x-a)^{2}. \label{equAh}
\end{equation}
Then $h$ is differentiable on $[a,b]$, and
\begin{equation}
h'(x)=f'(x)-\frac{f'(b)-f'(a)}{b-a}(x-a).
\end{equation}
It follows that $h'(a)=h'(b)=f'(a)$. Applying
Myers' theorem to $h$ gives $h(b)-h(c)=h'(c)(b-c)$ for some
$c\in (a,b)$. Rewriting $h$ and $h'$ in terms of $f$ gives the asserted
result.
\end{proof}
\begin{remark} \label{rmk2} \rm
It is easy to see that if $f'(a)=f'(b)$, then this result
reduces to Theorem \ref{thmD}. Furthermore, Theorem \ref{thm1} remains
valid if the function $h$ given by \eqref{equAh} is replaced by
\begin{equation}
h(x)=f(x)-\frac{1}{2}\frac{f'(b)-f'(a)}{b-a}(x-b)^{2}.
\end{equation}
This shows our that the function $h$ is not unique. So, we can find same
result by using different an auxiliary function $h$.
\end{remark}
The second goal of this paper is to prove a version of Myers' theorem for
holomorphic functions of one complex variable in the spirit of Evard and
Jafari \cite{EvardJafari}.
\begin{theorem} \label{thm3}
Let $f$ be a holomorphic function defined on an open convex subset $D_{f}$
of $\mathbb{C}$. Let $a$ and $b$ be two distinct points in $D_{f}$.
Then there exist $z_1,z_2\in (a,b)$ such that
\begin{equation}
\operatorname{Re}(f'(z_1))
=\frac{\langle b-a, f(b)-f(z_1)\rangle}{\langle b-a, b-z_1\rangle}-
\frac{1}{2}\frac{\operatorname{Re}(f'(b)-f'(a))}{b-a}(b-z_1)
\end{equation}
and
\begin{equation}
\operatorname{Im}(f'(z_2))=\frac{\langle b-a, -i[ f(b)-f(z_2)] \rangle }{
\langle b-a, b-z_2\rangle }-\frac{1}{2}\frac{\operatorname{Im}(f'(b)-f'(a))}{
b-a}(b-z_2).
\end{equation}
\end{theorem}
\begin{proof}
Let $u(z)=\operatorname{Re}(f(z))$ and $v(z)=\operatorname{Im}(f(z))$ for $z\in D_{f}$. We
now define the auxiliary function $\Phi :[0,1]\to \mathbb{R}$ by
\begin{equation}
\Phi (t)=\langle b-a, f(a+t(b-a))\rangle, \label{equAB}
\end{equation}
which is
\begin{equation}
\Phi (t)=\operatorname{Re}[(b-a)~u(a+t(b-a))]+\operatorname{Im}[(b-a)~v(a+t(b-a))]
\end{equation}
for every $t\in [ 0,1]$. Therefore, using the Cauchy-Riemann
equations, we obtain
\begin{align*}
\Phi '(t) &= \langle b-a,(b-a) f'(a+t(b-a))\rangle \\
&= \operatorname{Re}((b-a)^{2})\frac{\partial u(z)}{\partial x}
+\operatorname{Im}((b-a)^{2}) \frac{\partial u(z)}{\partial x} \\
&= | b-a| ^{2}\frac{\partial u(z)}{\partial x} \\
&= | b-a| ^{2}\operatorname{Re}(f'(z)).
\end{align*}
Applying Theorem \ref{thm1} to $\Phi $ on $[0,1]$, we obtain
\begin{equation}
(1-t_1)\Phi '(t_1)=\Phi (1)-\Phi (t_1)-\frac{1}{2}\frac{\Phi
'(1)-\Phi '(0)}{1-0}(1-t_1)^{2}
\end{equation}
for some $t_1\in (0,1)$. Thus
\begin{equation}
(1-t_1)| b-a| ^{2}\operatorname{Re}(f'(z_1))=\Phi
(1)-\Phi (t_1)-\frac{1}{2}[\Phi '(1)-\Phi '(0)](1-t_1)^{2}, \label{equAh1}
\end{equation}
where $z_1=a+t_1(b-a)$. Further, since $z_1=a+t_1(b-a)$ and $
t_1\in [ 0,1]$, we have $(1-t_1)| b-a|
^{2}=$. Hence the equation \eqref{equAh1} reduces to
\begin{equation}
\operatorname{Re}(f'(z_1))=\frac{\Phi (1)-\Phi (t_1)}{
(1-t_1)| b-a| ^{2}}-\frac{1}{2}\frac{\Phi '(1)-\Phi '(0)}{| b-a| ^{2}}(1-t_1).
\end{equation}
Using \eqref{equAB} and the fact that $z_1=a+t_1(b-a)$ in the above
equation, we obtain
\begin{equation}
\operatorname{Re}(f'(z_1))
=\frac{\langle b-a, f(b)-f(z_1)\rangle}{\langle b-a, b-z_1\rangle}-
\frac{1}{2}\frac{\operatorname{Re}(f'(b)-f'(a))}{b-a}(b-z_1).
\label{equAC}
\end{equation}
Letting $g=-if$, we have
\begin{equation}
\operatorname{Re}(g'(z))=\frac{\partial v(z)}{\partial x}=-\frac{\partial
u(z)}{\partial y}=\operatorname{Im}(f'(z)).
\end{equation}
Now, applying the first part to $g$, we obtain
\begin{equation}
\operatorname{Re}(g'(z_2))
=\frac{\langle b-a, g(b)-g(z_2)\rangle }{\langle b-a, b-z_2\rangle}-
\frac{1}{2}\frac{\operatorname{Re}(g'(b)-g'(a))}{b-a}(b-z_2)
\end{equation}
for some $z_2\in (a,b)$; i.e. $z_2=a+t_2(b-a)$ and $t_2\in [
0,1]$. By using \eqref{equAC}, the above equation yields
\begin{equation}
\operatorname{Im}(f'(z_2))=\frac{\langle b-a, -i[ f(b)-f(z_2)] \rangle}{
\langle b-a, b-z_2\rangle}
-\frac{1}{2}\frac{\operatorname{Im}(f'(b)-f'(a))}{b-a}(b-z_2).
\end{equation}
The proof is complete.
\end{proof}
It is easy to see that if $f'(a)=f'(b)$, then this result
reduces to the following complex version of Myers' theorem.
\begin{corollary} \label{coro4}
Let $f$ be a holomorphic function defined on an open convex subset $D_{f}$
of $\mathbb{C}$. Let $a$ and $b$ be two distinct points in $D_{f}$, and
$f'(a)=f'(b)$. Then there exist $z_1,z_2\in (a,b)$ such that
\begin{equation}
\operatorname{Re}(f'(z_1))
=\frac{\langle b-a, f(b)-f(z_1)\rangle}{\langle b-a, b-z_1\rangle}
\end{equation}
and
\begin{equation}
\operatorname{Im}(f'(z_2))=\frac{\langle b-a, -i[ f(b)-f(z_2)] \rangle}{
\langle b-a, b-z_2\rangle}.
\end{equation}
\end{corollary}
Returning to the example $f(z)=e^{z}-z$, $z_1$ and $z_2$ predicted by
Corollary \ref{coro4} have values $z_1\approx 1.78659i\in (0,2\pi i)$ and
$z_2\approx 3.94888i\in (0,2\pi i)$.
The third goal of this paper is to prove the equivalence of Rolle's and mean
value theorems in the complex plane, given by Evard and Jafari \cite{EvardJafari}.
\begin{theorem}[Complex Rolle's Theorem] \label{thmG}
Let $f$ be a holomorphic function defined on an open convex subset
$D_{f}$ of $\mathbb{C}$. Let $a,b\in D_{f}$ be such that
$f(a)=f(b)=0$ and $a\neq b$. Then there exist $z_1,z_2\in (a,b)$
such that $\operatorname{Re}(f'(z_1))=0$ and $\operatorname{Im}(f'(z_2))=0$.
\end{theorem}
\begin{theorem}[Complex Mean Value Theorem] \label{thmH}
Let $f$ be a holomorphic function defined on an open convex subset
$D_{f}$ of $\mathbb{C}$. Let $a$ and $b$ be two distinct points in
$D_{f}$. Then there exist $z_1,z_2\in (a,b)$ such that
$\operatorname{Re}(f'(z_1))=\operatorname{Re}\big( \frac{f(b)-f(a)}{b-a}\big)$
and $\operatorname{Im}(f'(z_2))=\operatorname{Im}\big( \frac{f(b)-f(a)}{b-a}\big) $.
\end{theorem}
\subsection*{Proof of the equivalence}
It is clear that Theorem \ref{thmG} must hold if Theorem \ref{thmH} does.
To show the converse, assume that
$f$ satisfy the conditions of Theorem \ref{thmH}. Then
\begin{equation} \label{equat3}
\begin{split}
g(z) &:=\frac{1}{a-b}\left|
\begin{matrix}
f(z) & f(a) & f(b) \\
z & a & b \\
1 & 1 & 1
\end{matrix}\right| \\
&= f(z)-f(a)\frac{z-b}{a-b}+f(b)\frac{z-a}{a-b}
\end{split}
\end{equation}
is also a holomorphic function for every $z\in D_{f}$. It is easy to see
that the function $g$ satisfies the condition $g(a)=g(b)=0$.
Hence, by Theorem \ref{thmG}, there exist $z_1,z_2\in (a,b)$
such that $\operatorname{Re}(g'(z_1))=0$ and $\operatorname{Im}(g'(z_2))=0$.
Thus, by \eqref{equat3}, we obtain
\begin{equation}
g'(z)=f'(z)-\frac{f(b)-f(a)}{b-a}
\end{equation}
for every $z\in D_{f}$. Hence,
\begin{gather}
0=\operatorname{Re}(g'(z_1))=\operatorname{Re}(f'(z_1))-\operatorname{Re}
\Big( \frac{f(b)-f(a)}{b-a}\Big) ,
\\
0=\operatorname{Im}(g'(z_2))=\operatorname{Im}(f'(z_2))-\operatorname{Im}
\Big( \frac{f(b)-f(a)}{b-a}\Big)
\end{gather}
which proves that Theorem \ref{thmG} implies Theorem \ref{thmH}.
Therefore, Theorems \ref{thmG} and \ref{thmH} are equivalent.
There are many ways to generalize the results of this paper due to the
papers \cite{FuriMartelli, McLeod, Rosenholtz, Samuelsson, Trahan} by using
the same method could be used here.
We skip further details in this regard.
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\end{document}